. Applied calculus; principles and applications . (a) J = Tr^ dA = J\ tt/ dp = ^ wd^32* 351 (1)(2)(3) (1) 352 INTEGRAL CALCULUS For a sector with angle 6, J = jdpdp = 4 _ ^1 _ 7rr^ (2)(3) (4) /. + /, = 27, = JFor a circular quadrant, 7, = 7For a semicircle, Ix — I (6) Since for the circle the moments of inertia about alldiameters are equal, 16 _ _ _ 7rr*^~2^ ~ 8 * 4. Ellipse: _ 1 ^2 1 (5)(6)(7) 7, = Jx dA = ^/x2 (a2 - a;2)ld(a; = ^. /x + 7, = ^(a2 + 62). (1)(2)(3). 187. Moment of Inertia of Compound Areas. — Sincethe moment of inertia is always positive, the moment ofinertia of an area about a
. Applied calculus; principles and applications . (a) J = Tr^ dA = J\ tt/ dp = ^ wd^32* 351 (1)(2)(3) (1) 352 INTEGRAL CALCULUS For a sector with angle 6, J = jdpdp = 4 _ ^1 _ 7rr^ (2)(3) (4) /. + /, = 27, = JFor a circular quadrant, 7, = 7For a semicircle, Ix — I (6) Since for the circle the moments of inertia about alldiameters are equal, 16 _ _ _ 7rr*^~2^ ~ 8 * 4. Ellipse: _ 1 ^2 1 (5)(6)(7) 7, = Jx dA = ^/x2 (a2 - a;2)ld(a; = ^. /x + 7, = ^(a2 + 62). (1)(2)(3). 187. Moment of Inertia of Compound Areas. — Sincethe moment of inertia is always positive, the moment ofinertia of an area about any axis is equal to the sum of the MOMENT OF INERTIA OF COMPOUND AREAS 353 moments of inertia, about that axis, of the parts into whichthe area may be divided. In some cases the area beingconsidered the difference of two areas, its moment of inertiawill be equal to the difference of the moments of inertia ofthe two areas. Example 1. — Find the momentof inertia of the T-section shownwith the dimensions on figure. (a) About the axis of Xthrough top of section. (6) About the axis Zo throughthe center of gravity. (a) 7. = i6¥ = 1(4-1) P + i(1 X 4^) = !+¥ = ¥- (ins.)*. 4 V -x-| 1 1 1—a:—^ f ■^ - h) forhollow square. h Example 3. — Hollow circular section: /
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